Understanding Fractions as a Form of Division
(Note that if you feel uncomfortable in working with variables you may want to check out the variables tutorial first. You may also find the scientific notation, order of operations, and exponents tutorials helpful as well.)
For students who make infrequent use of fractions, the process of working with fractions can be especially challenging. However, it will not be necessary to deal with all of the ways in which fractions can be manipulated. Instead, simply understanding that fractions can be a convenient representation of division is sufficient for your needs in this course, although the ability to reduce a fraction by canceling like terms in the numerator and denominator can be very helpful.
When you were first introduced to the concept of division back in elementary school, you were taught how to “do” long division through a process that you likely memorized and have since forgotten. That too will not be an issue in this course, thanks to the availability of calculators. You were also introduced to the symbol used in long division, \()\hspace{-4mu}\overline{\vphantom{\small)}\ \ \ \ \ \ }\), as well as the symbol \(\div\) meant to represent division within a sentence. Unfortunately neither of those representations is very useful within the context of equations. It is for this reason that the much more convenient representation as a fraction is used. To clarify, each of the following representations of “\(a\) divided by \(b\)” are equivalent:
$$b\,)\hspace{-4mu}\overline{\vphantom{\small)}\ a\ } \qquad a\div b \qquad \frac{a}{b} \qquad a/b.$$
Representing division with a fraction is more than just symbolic, it is truly the numerator of the fraction divided by the denominator. To see this, consider the fraction \(4/5\). As a fraction you should think along the lines of “there were originally five pieces of pie but now there are only \(4\) remaining, so the fraction of the pie remaining is 4/5ths (four-fifths).” However, that is the same thing as saying that “the amount of pie remaining is \(4/5\) or \(0.8\) of the original (\(80\%\)).” (If you are confused as to why \(0.8\) is the same thing as \(80\%\), see the percents tutorial.)
As another example, what is the result of dividing \(6\) by \(3\)? You probably recognize that the answer is \(2\) even without reaching for your calculator. However, fractions allow you to see this by “reducing” the fraction as follows:
$$\frac{6}{3} = \frac{3\times 2}{3} = \frac{3\times2}{3\times1} = \frac{3}{3}\times \frac{2}{1} = 1\times 2 = 2.$$
Recall that whenever you have a number (or variable) in the numerator (top) of the fraction that is multiplying all of the remaining numbers (or variables) in the numerator, and that same number (or variable) is also multiplying all of the remaining numbers (or variables) in the denominator (bottom) of the fraction, only then can the number can be canceled in both the numerator and the denominator.
Note that it is important to realize that you CANNOT cancel like terms in the numerator and the denominator if they are not multiplying all of the other terms. To see this consider the following example:
$$\frac{12}{2} = \frac{2+10}{2} {\color{red}\ne} \frac{1+10}{1} = 10.$$
Clearly \(\frac{12}{2}=6\), but by canceling terms incorrectly, you arrive at the value of \(10\). (You can interpret \(\ne\) as “not equal to.”) On the other hand, you CAN do the following:
$$\frac{12}{2} = \frac{2\times6}{2\times1} = \frac{2}{2}\times\frac{6}{1} = 6$$
The same kind of incorrect result occurs when you try to cancel the same variable in the numerator and the denominator if the variable is not multiplying every term in both the numerator and the denominator:
$$\frac{x + y}{x} {\color{red}\ne} \frac{1 + y}{1} {\color{red}\ne} y.$$
This last example is exactly the same as the first example in this paragraph if \(x=2\) and \(y=10\).
In many cases reducing the fraction in this way is not very helpful, and in fact can’t lead you to a single number as the answer without leaving the answer in some form of a fraction anyway. For example:
$$\frac{3.14159}{9.87654}$$
will not result in a simple result. In such a case, grab your calculator and carry out the division as
$$\frac{3.14159}{9.87654} = 3.14159÷9.87654=0.3180.$$
However, in other cases, especially those involving variables, reducing the equation can help to clarify what the equation is actually telling you. Consider this example:
$$\frac{6f}{3} = \frac{(3\times2)f}{3} = \frac{3}{3}\times \frac{2f}{1} = 1 \times 2f = 2f.$$
What \(\frac{6f}{3}\) is really saying is just “multiply \(f\) by \(2\); \(2f\) a significantly simpler statement that means exactly the same thing as the initial statement, “multiply \(f\) by \(6\) and then divide the whole thing by \(3\).”
When you “cancel” a term in the numerator (or variable) with the same number (or variable) in the denominator, what you are really doing is multiplying the numerator and denominator by the reciprocal of that number (or variable). Remember that you can multiply anything by \(1\) and not change that quantity’s value. Cleverly writing \(1\) as a fraction with the same value on top and on the bottom works very well for our purposes. To make the last example slightly more complicated (temporarily!),
$$\frac{6f}{3}=1×\frac{6f}{3}=\left(\frac{\frac{1}{3}}{\frac{1}{3}}\right)\times\frac{6f}{3}=\frac{\frac{1}{3}\times6f}{\frac{1}{3}\times3}=\frac{\frac{6}{3}f}{\frac{3}{3}}=\frac{2f}{1}=(2\times f)\div1=2f.$$
Recall that when you multiply two fractions together you must multiply the two numerators together and multiply the two denominators together which is what happened in going from the third step to the fourth step.
Evaluate the next expression, assuming that \(x=2\) and \(y=3\):
$$\frac{x^3}{xy^2}.$$
Before actually substituting numbers into the expression, hopefully you noticed that it can be reduced easily by canceling one common factor of \(x\) in both the numerator and the denominator which will reduce \(x^3=x\times x\times x\) in the numerator to \(x\times x=x^2\) while also reducing \(xy^2=x\times y\times y\) in the denominator to \(y\times y=y^2\), or
$$\frac{x^3}{xy^2}=\frac{x^2}{y^2}.$$.
Recalling that \(x=2\) and \(y=3\), you can now see that
$$\frac{x^3}{xy^2}=\frac{x^2}{y^2}=\frac{2^2}{3^2}=\frac{4}{9}=0.44444\ldots.$$
If your instructor wants you to fully evaluate some or all of the equations in the text, you will find yourself needing to divide numbers written in scientific notation. For example,
$$\frac{3.14159×10^9}{9.87654×10^2}.$$
Above it was pointed out that if you multiply two fractions together, you must multiply the two numerators and multiply the two denominators. This process can also be reversed; if you have an equation with two numbers multiplying one another in the numerator and two numbers multiplying one another in the denominator, then you can break up the single fraction into two fractions. Symbolically this means that
$$\frac{ab}{cd}=\frac{a}{c}\times\frac{b}{d}.$$
[By the way, since multiplication is commutative, meaning that the order of the variables (or numbers) doesn’t matter in multiplication, all of the following are equivalent:
$$\frac{ab}{cd}=\frac{a}{c}\times\frac{b}{d}=\frac{a}{d}\times\frac{b}{c}=\frac{ab}{c}\times\frac{1}{d}=a\times\frac{b}{cd}$$
and so on.]
Being able to factor fractions in this way comes in handy when working with scientific notation because scientific notation really is one number multiplying another number. In the example above,
$$\frac{3.14159\times10^9}{9.87654\times10^2}=\frac{3.14159}{9.87654}\times\frac{10^9}{10^2}$$
now becomes
$$\frac{3.14159}{9.87654}\times\frac{10^9}{10^2}=0.318086\times10^{9−2}=0.318086\times10^7=3.18086\times10^6.$$
The reduction of \(\frac{10^9}{10^2}=10^{9−2}=10^7\) comes from canceling \(2\) factors of \(10\) from both the numerator and the denominator, after all
$$10^9=10\times10\times10\times10\times10\times10\times10\times10\times10$$
and
$$10^2=10×10.$$
Eliminating \(2\) factors of \(10\) in the numerator with the \(2\) factors of \(10\) in the denominator corresponds to
$$\frac{10^9}{10^7}=10^{9−2}=10^7.$$
The middle expression is explicitly saying that two of the nine factors of \(10\) in the numerator are being canceled by the two factors of \(10\) in the denominator.
When you divide powers of ten (or any number or variable with the same base for that matter), you simply subtract the exponent in the numerator by the exponent in the denominator. We also saw this in the previous example:
$$\frac{x^3}{xy^2}=\frac{x^3}{x^1y^2}=\frac{x^{(3−1)}}{y^2}=\frac{x^2}{y^2}.$$
Consider one last example: For the expression below, let \(G=6.67\times10^{−11}\), \(m_1=100\), \(m_2=70\), and \(r=1.5\).
| \(\displaystyle G\frac{m_1\,m_2}{r^2}\) | \(=\) | \(\displaystyle\left(6.67\times10^{−11}\right)\times\left(\frac{100\times70}{1.5^2}\right)\) |
| \(=\) | \(\displaystyle\frac{\left(6.67\times10^{−11}\right)\times100\times70}{1.5^2}\) | |
| \(=\) | \(\displaystyle\frac{\left(6.67\times10^{−11}\right)\times\left(7\times10^3\right)}{2.25}\) | |
| \(=\) | \(\displaystyle\frac{\left(6.67\times7\right)\times\left(10^{−11}\times10^3\right)}{2.25}\) | |
| \(=\) | \(\displaystyle\frac{46.69\times10^{−8}}{2.25}\) | |
| \(=\) | \(\displaystyle\frac{46.69}{2.25}\times10^{−8}\) | |
| \(=\) | \(20.75\times10^{−8}\) | |
| \(=\) | \(2.075\times10^{−7}\). |
For the curious, you just calculated the attractive force of gravity (in newtons) between a \(100\) kg person and a \(70\) kg person separated by a distance of \(1.5\) m. This translates into a very tiny gravitational pull of \(2.075\times10^{-7}\) kg or \(4.5\times10^{−7}\) pounds (lbs), which is \(0.45\) micro pounds! “Big \(G\)” is the universal gravitational constant. [It is a fact of nature that everybody is attracted to everybody else (at least gravitationally).]
Quiz Yourself
Now that you have more familiarity with working with fractions, you should try some of the problems below to solidify your understanding. By the way, computers and many calculators write scientific notation in “computer-speak,” where \(10\) raised to an integer (either positive or negative) replaces \(\times 10^\text{exponent}\) with \({\rm E(exponent)}\) or \({\rm e(exponent)}\); in other words, \(3.14159 \times 10^9\) is represented by \(3.14159\text{E}9\). If your calculator has the \(\Large{\hat{ }}\) symbol, then you would enter the number as \(3.14159 \times 10{\Large{\hat{ }}}9\). If your calculator has a \(10^x\) button then you would enter the number as \(3.14159\times9\,\,10^x\). If you have having problems with entering numbers in scientific notation, read the manual or help screens, or ask your instructor.
- What is the value of \(\displaystyle\frac{932}{123}\)?
- Without using your calculator(!), what is the value of \(\displaystyle\frac{9\times10^{15}}{3×10^6}\)?
- Again, without using your calculator, what is the value of \(\displaystyle\frac{\left(3\times10^{12}\right)\times\left(4\times10^{−2}\right)}{6\times10^{−4}}\)?
- If \(a=6\), \(b=9\), and \(c=12\), evaluate \(\displaystyle\frac{a^2b^3}{abc^2}\). (Note that you should try to reduce this first before carrying out the calculations.)
Answers
- \(7.577\,24\)
- \(3\times10^9\)
- \(2\times10^{14}\)
- \(\displaystyle\frac{ab^2}{c^2}=\frac{27}{8} = 3.375\)